capacitorcurrentoperational-amplifier
I have been reading about op amps and came across this description of a differentiator variant using a capacitor.
The author explains that current can go through the capacitor when there is changing voltage, but I struggle to understand how the capacitor does not block the DC voltage source. Could anybody explain how there can be current in this situation?
Let's start our with a single op amp and work out way toward a class AB design. Consider the voltage follower.
simulate this circuit – Schematic created using CircuitLab
In this case \$V_{in}=V_{out}\$. Why? Negative feedback. Negative feedback forces the inverting pin's voltage to match the non-inverting pin. In other words, the op amp will do whatever it takes with its output to make \$V_{NI}=V_{INV}\$.
Let's take that a step further. We need more power to drive a low impedance speaker. Well, the average op amp will only have a few tens of milliamps of drive capability. That is where we add the power stage.
By tying the inverting pin to the output of the power stage, we have created a voltage follower. \$V_{in}=V_{out}\$, but now the circuit has the ability to deliver much more current than the op amp output ever could. Because the op amp has negative feedback, \$V_{NI}=V_{INV}\$. The cross over distortion is eliminated by the op amp doing whatever it takes to satisfy that relationship. As an exercise, build this up on a bread board, put a sine wave into \$V_{in}\$, and observe the output of the power stage and output of the op amp. The two will not look anything alike!
I think he got lucky- that hairball haywired circuit probably had some inductance that caused the oscillation. High currents only require a small layout issue to cause feedback voltage to show up- at the 750kHz that he reported and several amperes it wouldn't take much.
In answer to your question, yes there is a valid reason to use such a capacitor even though it generally reduces stability. It has to do with EMI sensitivity of op-amp front ends which can produce an apparent DC offset via nonlinearity of the front end response. Since low level signal generally require high gain, stability is less of an issue than it would be with a very low gain circuit.
Best Answer
That's pretty much the definition of a circuit that acts like a differentiator.
It does block DC - that's the whole point of a differentiator - DC is irrelevant.